Does the function below define an inner product? $$\langle (x, y), (z, t)\rangle = xz − yt$$
I know how to prove it given two vectors (e.g. $\langle(x,y),(z,t)\rangle$) demonstrating symmetry, linearity and positive-definiteness. But it's not clear to me how to deal with functions.
Should I find $u,v$ (e.g. $u=(x,y), z=(z,-t)$) and demonstrate the properties above?
Another example, a more complex one that would be hard to find $u,v$: $$\langle (x,y),(z,t)\rangle = 2xz -xt -yz +2yt$$
You need to demonstrate or refute the properties of an inner product for $$ \langle (x,y) , (z,t)\rangle = xz -yt$$
If you suspect that it is not an inner product then yes, you need to find suitable vectors that violate one of the properties of an inner product.
If $(x,y), (z,t)$ are vectors in $\mathbb R^2$ then pick for example $(x,y) = (z,t) = (1,2)$ and note that $\langle (1,2) , (1,2) \rangle = 1-4 < 0$ so that positive-definiteness is violated and it cannot be an inner product.
For the other example you give you need to proceed in the same manner: first establish whether you want to prove or refute it and then either try to find counterexamples or try to find a proof.
For a proof you need to verify the following four properties of inner products:
(i) Symmetry: $\langle x,y\rangle = \langle y,x\rangle$
(ii) Linearity: $\langle \alpha x + \beta x',y\rangle = |\alpha|\langle x ,y\rangle + |\beta| \langle x',y\rangle$
(iii) $\langle x,x\rangle \ge 0$
(iv) $\langle x,x \rangle = 0$ iff $x=0$
Example (symmetry):
$\langle (x,y),(z,t)\rangle = 2xz -xt -yz +2yt$
$\langle (z,t), (x,y) \rangle = 2xz - yz -xt + 2yt$